Mass Transfer

Section # 9Mass Transfer 2Sheet#2 , Problem#1Givens:Air-water vapor mixtureT=65oC, Tw=35oC, PT = 1 atmThe radiation coefficient can be considered negligibleAntoine constants for waterThe steam tableRequired:The humidity of the air without using the chart

Sheet#2 , Problem#1Solutionw - ww = - hg [ 1 / Ky Mg v ] (1) t - twFor air-water system, the term (hg / Ky Mg) = 0.26 BTU / lb oF = 1.09 j / g oCFrom the steam table, table 1, you will find that at Tw = 350C

= 2418.8 j/g , saturation pressure = 0.056 atmNote: you can also calculate the saturation pressure using Antoine equation at Tw=35oC.Ww = Ps * MV = 0.056 * 18 = 0.0368 kg water P- Ps Mg 1 - 0.056 29 kg dry airSub. in (1) get w = 0.0229

Sheet#2 , Problem#2Givens:Air-Toluene vapor mixtureT=60oC, w = 0.05 kg vapor / kg airD= 9.2 * 10-6 m2/sec, = 1.95 * 10-5 kg/ m sec, k = 0.0277 W / m KA table for toluene properties hc = Sc 0.567 = Le 0.567 Ky Cs prRequired:The wet bulb temperature of the mixtureThe adiabatic saturation temperature of the mixture

Sheet#2 , Problem#2SolutionSc = , Pr = * Cs * D k = PM = 1 * 29.1 = 1.065 g / L Sc = 1.99 RT 0.082 * (60+273)Cs = Cg + w Cv = 1.005 + (0.05* 1.82) = 1.096 j / g K Pr = 0.772But, hc = Sc 0.567 = 1.71 (hc / Ky ) = 1.71 * 1.096 = 1.875 Ky Cs Pr w-ww = - hg [ 1 / Ky Mg v ] 0.05-ww = - 1.875 (1) t-tw 60 - tw vNow, a trial and error procedure is needed. We will perform the following: a) Assume tw b) Get v , and the toluenes saturation pressure from the toluenes table c) Calculate ww using the got saturation pressure d) Check that the L.H.S and R.H.S of eqn. (1) has the same value. 5Sheet#2 , Problem#21st assumption: Tw = 30oC v = 409.56 kj /kg , and Ps = 0.04972 barWw = Ps * MV = 0.04972 * 92 = 0.166 kg toluene P- Ps Mg 1 -0.04972 29 kg dry airL.H.S of eqn. (1) = -0.003866, R.H.S of eqn. (1) = -0.0045772nd assumption: Tw = 35oC v = 406.66 kj /kg , and Ps = 0.06292 barWw = Ps * MV = 0.06292 * 92 = 0.213 kg toluene P- Ps Mg 1 -0.06292 29 kg dry airL.H.S of eqn. (1) = -0.00652, R.H.S of eqn. (1) = -0.00461We can get the true wet bulb temp. using interpolation: Tw 30 = 0 (0.004577-0.003866) 35 -30 (0.00461 0.00652) ((0.004577-0.003866)

Get Tw = 31.356 oCSheet#2 , Problem#2w-ws = - Cs / s 0.05-ws = - 1.096 (2) t-ts 60 - ts sYou can use the same technique performed in estimating the wet-bulb temperature here to estimate the adiabatic saturation temperature. After trial and error, you can get that: Ts = 27oCSheet#2 , Problem#7Givens:Dehumidification tower, counter-currentLiquid rate = 1000 lb/hr ft2 , TL2=60oF , TL1=90oFTv1=100oF, % Humidity1=90%Gas rate = 27000 std ft3/hr ft2hLa / kya = 200Required:The outlet air conditions

Sheet#2 , Problem#7 SolutionHv2 - Hv1 = (L avg CL ) / S Hv2 - 61 = 1000*1(1) TL2 TL1 V / S 60 90 GWhere, Hv1 was calculated by knowing 2 information about the entering gas.Since gas rate = 27000 std ft3/ hr ft2, we want to convert it into lb/ hr ft2 to be able to sub. In the above operating line equation.For ideal gases : PV = nRT 14.7 * 27000 = n*10.72*492 get n1 = 75.25 lbmole / hr ft2 Note: (std means that T = 0oC, and pressure = 1 atm)n = 75.25 lbmole air / hr ft2 G1 = 75.25 * 29 = 2182.25 lb air / hr ft2G1= G* (1+w1) G = G1 / (1+0.041) = 2096.3 lb dry air / hr ft2Sub. in (1) Get Hv2 = 46.689 BTU / lb dry airAs TL2=45oC, and TL1=29oC, then the temperature range that we will take to draw the equilibrium curve will be (50, 25 oC)Steps: a) Draw the equilibrium curve on a graph paper b) From Tw1 and T1, get Hv1 from the chart c) As we have TL1 , and Hv1 , then we can locate point (1) on the operating line d) Draw a locus for TL2 e) Match point (1) with the (intersection between the equilibrium curve and the locus of TL2 ) to get the pinch point.Sheet#2 , Problem#7The required was the outlet air conditions. The humid enthalpy is not enough.We will apply Mickleys method so as to get Tv2To apply Mickleys method, we should have the slope of the tie-lines.It is given that hLa / kya = 200. This is a very large value for the slope. In this case, the slope is vertical.After performing Mickleys method Get Tv2 = 87oFYou will find that the point describing the outlet air conditions here lies on the eqm. curve % RH2 = 100 % , and Tv2=Tw2

Sheet#2 , Problem#7

Sheet#2 , Problem#7

Sheet#2 , Problem#7

Sheet#2 , Problem#8Givens:Dehumidification tower, counter-currentLiquid rate = 900 lb/hr ft2 , TL2 = 82oF, TL1 = 100oFTv1=125oF, Tw1= 111oF Tv2=96oF , Tw2=95oFZ = 8 ftRequired:hca, hLa, and kyaIn the same tower, if the water inlet temperature, air and water flow rates would not change, but the inlet gas conditions become 140oF dry bulb, and 111oF wet bulb. Determine the outlet air conditions. You can use the coefficients calculated from part (a)

Sheet#2 , Problem#8 SolutionWhen Z, TL1, TL2, Tv1, Tv2, additional information on the inlet air, and an additional information about the exiting air are given, Reverse Mickley technique can be used to be able to calculate hca, hLa, and kya.How to perform Reverse Mickleys method? a) By knowing Tv1 and Hv1,you can locate the point representing the entering gas. You can name it (1). b) In this case, you can also represent the exiting gas by knowing Tv2 and Hv2 where the point representing it is point (2). c) Assume a tie-line slope, then from point (1), apply Mickleys technique as we have done before d) Repeat the steps till you reach the locus of Hv2. If the point got on this locus is (point 2), then the assumed tie line slope is true. If not, reassume the slope.Sheet#2 , Problem#8

Sheet#2 , Problem#81st trial: a) Assume (- hLa / kya) = -1.5 b) From point (1), start Mickley c) Continue Mickleys method till you reach the locus of Hv2. If you cut the locus in Tv2, then the assumed slope is ok. d) You will find that this slope will cut the Hv2 locus in 95oF (not 96oF)

2nd trial: a) Assume (- hLa / kya) = -11 b) From point (1), start Mickley c) Continue Mickleys method till you reach the locus of Hv2. If you cut the locus in Tv2, then the assumed slope is ok. d) You will find that this slope will cut the Hv2 locus in 96.5oF (not 96oF)

You will find that the true tie-line slope is -4.66 BTU / lb oF

Sheet#2 , Problem#8 SolutionHv2 - Hv1 = (L avg CL ) / S 57 - 89 = 900*1(1) TL2 TL1 V / S 82 100 G Where, Hv1 and HV2 were calculated by knowing 2 information about the entering and exiting gas respectively.Get G = 506.25 lb / hr ft2 Z = HTG * NTG 8 = HTG*NTG (2) Where: Hv2 NTG = d Hv ( It can be calculated as we know everything) Hv1 Hi HvHTG = (G / kya) kya is unknown

Sheet#2 , Problem#8 Hv2 NTG = d Hv We need the tie-line equation Hv1 Hi HvHv Hi = - hL a = - 4.66 BTU / lb o F TL T I kya

Get NTG = 2.1145Sub. in (2) get HTG = 3.7833 ft 3.7833 = 506.25 / kya kya = 133.81 lb / hr ft3But - hL a = - 4.66 BTU / lb o F hLa = 623.56 lb / hr ft2 kya

Hv5765738189Hi43.5505765751 / (Hi HV)0.0740.06660.06250.06250.0714Sheet#2 , Problem#8For air-water system, r = 1 where: r = psychrometric ratio = hca / kya Cs (3)

Cs = Cg + wavg Cv = 0.24 + wavg (0.45)wavg = (w1 +w2) / 2 = (0.074 + 0.036) / 2 = 0.055 Cs = 0.2652 BTU lb oFSub. in (3) by r = 1 hca = 35.48 BTU / hr ft3oF

Sheet#2 , Problem#8 Solutionb) TL2, Lavg, G, Z, kya, hLa are the same(1)Tw1 is also the same Hv1 is also the same = 89 BTU / lb dry air HTG = (G / kya) From (1), HTG is the same = 3.7833 ftSince Z and HTG arent changed, NTG will not also be changed Hv2 Therefore, NTG = d Hv = 2.1145 Hv1 Hi HvSince the tie-line slope and Hv1 are the same, then Hv2 will not be changed. Hv2 = 57 BTU / lb dry airHv2 - Hv1 = (L avg CL ) / S 57 - 89 = 900*1 TL1 = 100oF TL2 TL1 V / S 82 TL1 506.25Sheet#2 , Problem#8Therefore, the operating line will not change. Of course, the equilibrium curve also will not change.The required was the outlet air conditions. The humid enthalpy (Hv2) is not enough.We will apply Mickleys method so as to get Tv2In this case, point (1) is (Tv1=140oF, Hv1= 89 BTU / lb dry air)The tie-line slope is the same as got in the 1st requirement (= - 4.66)After performing Mickley, get Tv2 = 97oFFrom the chart, at Tv2 = 97oF, and Hv2 = 57 BTU/lb dry air w2 = 0.0385Sheet#2 , Problem#8